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%TCIDATA{Created=Mon Jul 19 06:09:18 2010}
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\begin{document}

\title{On the measure of quality of BS formula: theoretical and empirical analysis}
\author{Diep Nguyen, Serguey Khovansky}
\date{\today}
\maketitle

\abstract{\ }

\begin{center}
to be added
\end{center}

\raggedright

\section{Introduction}
As financial derivative products become popular to the financial markets, how to price them becomes of interest to market participants as well as scholars. One of the most popular models for pricing options and related financial derivatives is the Black-Scholes formula, which has been widely used and credited for its practical relevance. \newline

However, as shown by Rubinstein ~\cite{Rubinstein1994} in his paper Implied Binomial Tree, the Black-Scholes formula has become 'increasingly unreliable over time'. An evidence for this unreliability is the volatility smile, which reflects deviation from a theoretically horizontal line of an option's implied volitility as a function of its strike price. As shown by John Hull  ~\cite{Hull2009}, the implied volatility, instead of being constant with respect to all strike prices, is lower for at-the-money option than for in-the-money and out-of-the-money ones. This forms a 'smile' shape when plotted against strike prices in the period of the stock crash in October 1987. Despite being one of the most remarkable findings in economics, the Black-Scholes formula does not always predict option price the same as the market does. This leads to the need to measure the accuracy of this model. \newline

Several methods have been employed in order to test if the Black-Scholes formula is an ultimate model to price options. The list includes put-call parity, comparing otherwise identical options of different maturities, comparing today's and tomorrow's implied volatilities of the same option, and comparing otherwise identical options of different strike prices. To contribute to the existing methods, we further analyze the minimax statistic method used in Rubinstein, 1994 and develop a new quality of fit measurement, which we will call the minisum statistic. Our interest in the two measurements comes from the fact that they are indepedent of implied volatilities and that they can be calculated analytically. The first makes the 'true' implied volatility irrelevant to the measurement and the second makes it practical for getting the measurement through computer programs. \newline

We will also perform empirical tests on the quality of fit of the Black-Scholes formula using the two measurements. The test will be performed on two options: the option on Standard $\&$ Poor's Depositary Receipts (SPY), listed on the New York Stock Exchange, and the Hang Seng Index Options, listed on the Hong Kong Exchanges. Rubinstein made an assessment on the performance of the Black-Scholes formula across different time periods. Our work extends the assessment to measuring the formula's accuracy across different markets whose levels of efficiency can be different.\newline

The paper is organized as follow. First we will provide analytical and computational analyses on the two measurements introduced above: the minimax statistic and the minisum statistic. Then we will discuss the empirical findings through assessment of the Black-Scholes formula's accuracy on the US market and the Hong Kong market using the two measurements analyzed in the previous part.

\section{A Measure of Quality of BS formula}

The quality of fit of the BS formula could be assessed by examining the
implied volatility, which according to the model assumptions must be a
constant. According to Black-Scholes model, the implied volatility must be not only independent on moneyness/maturity and but also equal the market volatility. In other words, not only the slope of the implied volatility must be zero according to the BS assumptions, but also it must be equal to the market volatility. However, in reality there is varying dependence of implied volatility on the option moneyness (i.e. smile or smirk) and maturity, which invalidates this postulate. To answer the question on how far the actual market prices deviate from the theoretical calculations we could explore the empirical properties of the implied volatility curve, by converting it to some index, which should have at least two properties 1) increase with the discrepancy between theoretical and actual prices, and 2) be zero (or some other known constant) if discrepancy is none, regardless of the actual level of volatilities. \newline

Rubinstein, 1994, suggested a minimax statistic, which sets a lower bound on the error of the Black-Scholes formula, i.e. however large the 'true' implied volatility is, the error of the Black-Scholes formula will be at least this minimax statistic value. The minimax statistic is used in literature as an unbiased mean to place bounds on the price of assets.  Pyo, 2009 ~\cite{Pyo2009} used the minimax deviation method, in which the maximum deviation is minimized, to set bounds on option price with regard to the level of risk averse of the agency. \newline

In his paper, Rubinstein provided a numerical description on how to get the minimax statistic. It can be constructed by comparing two options different only by their strike prices. For each option, the dollar error, which is the absolute difference between the price of an option and it market price, is calculated for a certain assumed volatility. A funtion of the dollar error with respect to assumed volatility is then built for each option as assumed volatility spans from 0 to infinity, which is the domain for the assumed volatility. At each volatility, we choose the greater of the two dollar errors of the two options. A function mapping each assumed volatility with the greater of the two dollar errors can be built. This will make up the 'max' part of the minimax statistic. The minimum value of this 'max' part is actually the minimax statistic as used by Rubinstein. On how to calculate the statistic, he stated that the volatility at which the minimax statistic is reached lies in between the two implied volatilities of the two options and makes the two dollar errors equal. We provide detailed theoretical examination of the minimax statistic which allows to compute their values analytically. This work extends the Rubinstein's results whose research was limited by numerical analysis only. \newline

Apart from that measure we introduce another criterion: minisum, which is proportional to the minimum of average dollar error. Both measures examine the extent to which the Black-Scholes formula deviates from the market price, without making any assumption about the implied
volatility. In addition, both measures equal zero when the BS formula
perfectly fit the actual prices without violating all underlying
assumptions. It should be noted that the direct analysis of the implied vol
can not be used to study fit because it depends on the level of market
volatility. High level of the implied vol can be caused by either strong
discrepancy between theory and practice or by the high market volatility.
The suggested measures bypass this issue by exploring the slopes on the
implied volatilities rather then their absolute level.\newline


\subsection{Minimum Average Dollar Error: Minisum function}

In this section we introduce a measure of the quality of
the Black-Scholes formula: the minimum average dollar error. If we take two
options of different strike prices with everything else equal, then the
minimum average dollar error (minisum) for these two options will be the
minimum of the average of their dollar errors. The idea behind this measure
is that the average of the dollar errors is at least this value, regardless
of the true underlying volatility. \newline


Next we will formulate this measure. Let the function $C^{BS}(X,\sigma )$
represent a Black-Scholes option price, given the strike $X$ and volatility $%
\sigma $. It also depends on a set of other parameters $\theta $ $=\{S,\tau
,r,\delta \}$, where $S$ is the underlying stock price, $\tau $ is the time
to expiration, $r$ is the risk free rate, and $\delta $ is the dividend
rate. To streamline the notation the dependence of $C^{BS}(X,\sigma )$ on $%
\theta $ is only implicitly assumed. The market option price is denoted by$%
C^{mkt}(X)$ which explicitly depends on the strike and implicitly on the
parameter set $\theta $. \textbf{To access the quality of the Black-Scholes}
formula we introduce a function 
\[
f(\sigma )=\left| C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )\right| +\left|
C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )\right| , 
\]
where both options are written on the same underlying asset. To evaluate the fit of the BS formula we use minisum function equal $\min_{\sigma }f(\sigma ).$ This is a meaningful measure of fitness for the Black-Scholes formula as it 1) increases if the dollar error of any of the two option increases and 2) is zero if the market is pricing options according to the Black-Scholes formula. \newline


 Because the average dollar can be obtained by taking the sum of dollar
errors and divide by a factor of 2, Theorem 1 below, which examines behavior
of the sum of dollar errors, can be applied to understand the properties of
the minimum average dollar error. The theorem provides the exact location of the minimum of $f(\sigma )$ through exploring the
possible minimums of the function $f(\sigma )$.

\begin{theorem}
\label{Theorem 1}Assume that $X_{1}<X_{2}$. Denote by $\sigma _{i}$ the implied volatility of the
option $C^{mkt}(X_{i})$, i.e. $C^{mkt}(X_{i})$ $=$ $C^{BS}(X_{i},\sigma
_{i}) $. Let $\varphi \equiv \frac{1}{\Delta t}\ln \frac{X_{1}X_{2}}{S_{t}^{2}%
}-2r$, where $\Delta t$ is the time to maturity, and let $h(\sigma)$ $\equiv$ $C^{BS}(X_{1},\sigma)-C^{BS}(X_{2},\sigma)$. Then the function $%
f(\sigma )=$ $\left| C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )\right| $ $+$ $%
\left| C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )\right| $ reaches its minimum at
the point $\sigma ^{*}$, where 

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{*}=\sigma _{2}$ and $\min $ $f(\sigma )$= $f(\sigma _{2})$= $%
\left| C^{BS}(X_{1},\sigma _{2})-C^{mkt}(X_{1})\right| $ & if $\varphi \leq 0
$ \\ 
$\sigma ^{*}=\sigma _{2}$ and $\min $ $f(\sigma )$= $f(\sigma _{2})$=$\left|
C^{BS}(X_{1},\sigma _{2})-C^{mkt}(X_{1})\right| $ & if $\varphi >0$ and $0<%
\sqrt{\varphi }<\min (\sigma _{1},\sigma _{2})$ \\ 
$\sigma ^{*}=\arg \min \left( h(\sigma _{1}),h(\sigma _{2})\right) $ and $%
\min $ $f(\sigma )$= $\min $ $(f(\sigma _{1}),f(\sigma _{2}))$ & if $\varphi
>0$ and $\sigma _{1}\leq \sqrt{\varphi }\leq \sigma _{2}$\\
$\sigma ^{*}=\sqrt{\varphi }$ and $\min $ $f(\sigma )$= $f(\sqrt{\varphi })$
& if $\varphi >0$ and $\sigma _{2}\leq \sqrt{\varphi }\leq \sigma _{1}$ \\ 
$\sigma ^{*}=\sigma _{1}$ and $\min $ $f(\sigma )$= $f(\sigma _{1})$=$\left|
C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma _{1})\right| $ & if $\varphi >0$ and $%
\sqrt{\varphi }>\max (\sigma _{1},\sigma _{2})$
\end{tabular}
\right. 
\]
\end{theorem}



Note that the expression $\sigma ^{*}=\arg \min \left(\text{ }f(\sigma _{1}),\text{ }f(\sigma _{2})\right) $ is not a function
since e.g. in the case when $f(\sigma _{1})=f(\sigma _{2})$ the $\sigma ^{*}$
can be take both values $\sigma _{1},$ and $\sigma _{2}.$ However, in the
applied part of the research we are interested in the minisum, i.e. min
value of the function $f()$ so this lack of functionality is not an issue
for this study.\newline

{\bfseries Intuition for the proof-} Our task is to find the global minimum for $%
f(\sigma )$, sum of the two dollar errors. However, the function $f(\sigma )$
is not differentiable everywhere. Hence, we need to find and compare the
boundaries, the stationeries, and the non-differentiable points to find the
global minimum for $f(\sigma )$ on its domain. To do this, we split the
domain of $f(\sigma )$, the interval $(0,\infty )$, into subintervals where
the differentiability and the value of $f(\sigma )$ can be determined. By
comparing the values of $f(\sigma )$ at the boundaries, the stationaries,
and the non-differentiable points, we can find the minimum of $f(\sigma )$
on its domain.

\subsection{The minimax statistics}

This subsection discusses minimax statistic, which is another
measure of the quality of the Black-Scholes formula employed by Rubinstein
(1994). For two otherwise identical options with different strike prices, the minimax statistic is the global minimum value of the greater of the two dollar errors at each assumed volatility. In other words, we get the greater of the two dollar errors for each assumed volatility. The minimax statistic is the minimum of those errors. The minimax statistic is considered a lower bound placed on the error of the Black-Scholes formula, i.e. for the two options, the Black-Scholes formula has at least this error. \newline

The minimax statistic can be formulated as follow. Let the function $D(X, \sigma)$ be the dollar error for an option of strike price $X$ at assumed volatility $\sigma$. $S$=$\{X_{1},X_{2}\}$ is the domain of the strike price, which receives the value of either $X_{1}$ or $X_{2}$. $V$=$(0, \infty)$ is the domain of the assumed volatility. Then the minimax statistic can be expressed as:

\[ Minimax Statistic = \min_{\sigma \in V} \max_{X \in S} D(X,\sigma)\]

The minimax statistic is illustrated in Figure \ref{Graph1}. The red dash line shows the max of the two dollar errors. Minimax statistic is the intersection of Dollar Error $X_{1}$ and Dollar Error $X_{2}$ and it is the minimum point of
the red dash line. 

\begin{figure}[h]
  \centering
        \includegraphics[scale=0.6]{GraphMinimax}
        \caption{Minimax Statistic}
    \label{Graph1}
\end{figure}


As stated in Rubinstein, 1994, ~\cite{Rubinstein1994} an important property of minimax statistic is equality of dollar errors at two strikes, i.e. $|C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma^{mm})|$ =$|C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma ^{mm})|$. We prove this statement in the Theorem \ref{Theorem 2}

\begin{theorem}
{\label{Theorem 2}} The argument of the minimax statistic is reached at a  unique assumed volatility $\sigma ^{mm}$ within the interval $\left[ \sigma
_{1},\sigma _{2}\right] $ such that the two dollar errors at this assumed volatility are equal, that is $|C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma ^{mm})|$ $=$ $|C^{mkt}(X_{2})$ - $C^{BS}(X_{2},\sigma ^{mm})|$.\newline
\end{theorem}

To prove Theorem \ref{Theorem 2}, we first show that there exists one and
only one argument $\sigma ^{mm}$ in $\left[ \sigma _{1},\sigma _{2}\right] $
that makes the two dollar errors equal. Then we prove that the dollar error
at $\sigma ^{mm}$ is the minimax statistic on the domain of $\sigma $, that
is minimax function at other volatilities exceed that at the $\sigma ^{mm}$.


\section{Empirical Investigation}

\textit{\ SK: Here we report our empirical investigation of the formulas
defined earlier}

\subsection{Data}
We use daily price data for the options writen on the S$\&$P 500 Index and the Hang Seng Index in this study. The first one is the option writen on the Standard $\&$ Poor's Depositary Receipts (symbol SPY), which tracks the S$\&$P 500 index. The Index is one of the most followed stock indices in the US and is considered as the bellweather for the US economy. Option prices are downloaded daily from the Yahoo!Finance website.\newline

The Hang Seng Index option is an European option writen on the Hang Seng Index ETF (stock code 2833), which is the benchmark for the Hong Kong stock market. The prices for this option can be downloaded daily from the Hong Kong Exchanges website. 

\subsection{Results of US data}

\subsection{Results of Hong Kong data}

\subsection{Results of some other data}

\section{Conclusion}

\section{References}
\bibliography{References}


\section{Appendix}

\subsection{Theorem Proofs}

\textbf{Theorem \ref{Theorem 1}. }Assume that $X_{1}<X_{2}$. Denote by $\sigma _{i}$ the implied volatility of the
option $C^{mkt}(X_{i})$, i.e. $C^{mkt}(X_{i})$ $=$ $C^{BS}(X_{i},\sigma
_{i}) $. Let $\varphi \equiv \frac{1}{\Delta t}\ln \frac{X_{1}X_{2}}{S_{t}^{2}%
}-2r$, where $\Delta t$ is the time to maturity, and let $h(\sigma)$ $\equiv$ $C^{BS}(X_{1},\sigma)-C^{BS}(X_{2},\sigma)$. Then the function $%
f(\sigma )=$ $\left| C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )\right| $ $+$ $%
\left| C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )\right| $ reaches its minimum at
the point $\sigma ^{*}$, where 

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{*}=\sigma _{2}$ and $\min $ $f(\sigma )$= $f(\sigma _{2})$= $%
\left| C^{BS}(X_{1},\sigma _{2})-C^{mkt}(X_{1})\right| $ & if $\varphi \leq 0
$ \\ 
$\sigma ^{*}=\sigma _{2}$ and $\min $ $f(\sigma )$= $f(\sigma _{2})$=$\left|
C^{BS}(X_{1},\sigma _{2})-C^{mkt}(X_{1})\right| $ & if $\varphi >0$ and $0<%
\sqrt{\varphi }<\min (\sigma _{1},\sigma _{2})$ \\ 
$\sigma ^{*}=\arg \min \left( h(\sigma _{1}),h(\sigma _{2})\right) $ and $%
\min $ $f(\sigma )$= $\min $ $(f(\sigma _{1}),f(\sigma _{2}))$ & if $\varphi
>0$ and $\sigma _{1}\leq \sqrt{\varphi }\leq \sigma _{2}$\\
$\sigma ^{*}=\sqrt{\varphi }$ and $\min $ $f(\sigma )$= $f(\sqrt{\varphi })$
& if $\varphi >0$ and $\sigma _{2}\leq \sqrt{\varphi }\leq \sigma _{1}$ \\ 
$\sigma ^{*}=\sigma _{1}$ and $\min $ $f(\sigma )$= $f(\sigma _{1})$=$\left|
C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma _{1})\right| $ & if $\varphi >0$ and $%
\sqrt{\varphi }>\max (\sigma _{1},\sigma _{2})$
\end{tabular}
\right. 
\]


\textit{Note: We call the expression $\min f(\sigma)$ minisum (dollar error) statistic.}



\subparagraph{Proof of theorem \ref{Theorem 1}.} 

Depending on the form of volatility smile, the implied volatility parameter of the function $f(\sigma ),$ i.e. $\sigma _{1}$ can be less, equal, or greater than another (implied volatility) parameter $\sigma _{2}$. For example, as discussed by John Hull, ~\cite{Hull2009}, the volatility smile curve during the stock market crash in October 1987 shows that implied volatility of at-the-money option is significantly lower than those of of out-of-the-money and in-the-money options, whose have higher and lower strike prices, respectively, than at-the-money option. However, before this stock crash, implied volatility had been rather independent of strike prices.  Correspondingly, we have to examine these three cases separately, that is when $\sigma_{1} < \sigma_{2}$, $\sigma_{1}=\sigma_{2}$, and $\sigma_{1} >\sigma_{2}$.\newline

For each case, the function $f(\sigma )$ is not differentiable at every point of its domain, so the standard method of finding minimum is not applicable. Instead we split the domain of the function into non-overlapped intervals, explore its behavior on each one and find the minimum. 

\smallskip

\begin{itemize}
\item  \underline{Case \#1: $\sigma _{1}$ $<\sigma _{2}$ }- To proceed, we
split the domain for the argument $\sigma $ of the function $f(\sigma )$
into 3 intervals: $[0,\sigma _{1})$ $\cup $ $\left[ \sigma _{1},\sigma
_{2}\right] $ $\cup $ $\left( \sigma _{2},\infty \right) .$

\begin{itemize}
\item  For $\sigma $ $\in $ $[0,\sigma _{1})$, the function $f(\sigma )$ $%
=C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )+C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )$
is decreasing in $\sigma $ because $\frac{\partial C^{BS}(X,\sigma )}{%
\partial \sigma }$ is positive for every strike $X$ (see Lemma \ref{Lemma 1}%
). \textit{\ }. Also, $\displaystyle \lim_{\sigma \to \sigma_{1}^{-}}f(\sigma)$=$C^{mkt}(X_{2})-C^{BS}(X_{2}, \sigma_{1})$. Hence $f(\sigma)$ in this interval will be greater than the limit value $C^{mkt}(X_{2})-C^{BS}(X_{2}, \sigma_{1})$.

\item  For $\sigma $ $\in $ $\left[ \sigma _{1},\sigma _{2}\right] $, the $%
f(\sigma )$ $=$ $-C^{mkt}(X_{1})+C^{BS}(X_{1},\sigma
)+C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )$ $=$ $-C^{mkt}(X_{1})$ $%
+C^{mkt}(X_{2})$ $+$ $h(\sigma )$, where $h(\sigma )=C^{BS}(X_{1},\sigma
)-C^{BS}(X_{2},\sigma ).$ Applying the Lemma \ref{Lemma2} we find that the
arg minimum of $f(\sigma )$ corresponds to the argmin of $h(\sigma )$ as
defined and explained in the Lemma \ref{Lemma2}.

\item  For $\sigma $ $\in \left( \sigma _{2},\infty \right) $, the function $%
f(\sigma )=$ $-C^{mkt}(X_{1})+C^{BS}(X_{1},\sigma
)-C^{mkt}(X_{2})+C^{BS}(X_{2},\sigma )$ is increasing in $\sigma $ because $%
\frac{\partial C^{BS}(X,\sigma )}{\partial \sigma }$ is positive for every $X
$ (see Lemma \ref{Lemma 1}). \textit{\ }.  Also, $\displaystyle \lim_{\sigma \to \sigma_{2}^{+}}f(\sigma)$=$-C^{mkt}(X_{1})+C^{BS}(X_{1}, \sigma_{2})$. Hence $f(\sigma)$ in this interval will be greater than the limit value $-C^{mkt}(X_{1})+C^{BS}(X_{1}, \sigma_{2})$.
\end{itemize}
\end{itemize}


The behavior of $f(\sigma)$ when $\sigma_{1}<\sigma_{2}$ is summarized in the table below:
\begin{table}[ht]
\caption{Examining $f(\sigma)$ in the case $\sigma_{1}<\sigma_{2}$} % title of Table
\centering % used for centering table
\begin{tabular}{c c c c} % centered columns (4 columns)
\hline\hline %inserts double horizontal lines
	& $0 < \sigma < \sigma_{1}$& $\sigma_{1} \leq \sigma \leq \sigma_{2}$ & $\sigma > \sigma_{2}$ \\ [0.5ex] 
\hline % inserts single horizontal line
	$f(\sigma)$ & Decreasing & Depends on behavior of $h(\sigma)$ & Increasing \\
\hline % inserts single horizontal line
	$\min f(\sigma)$ & Greater than $f(\sigma_{1})$ & Is either $f(\sigma_{1})$ or $f(\sigma_{2})$ & Greater than $f(\sigma_{2})$\\
\hline
\end{tabular}
\label{table1} % is used to refer this table in the text
\end{table}

\begin{description}
\item  The function $f(\sigma )$ is decreasing on $[0,\sigma _{1})$ and
increasing on $\left( \sigma _{2},\infty \right) $, hence its argmin is in
the $\left[ \sigma _{1},\sigma _{2}\right] $ interval. Applying the Lemma 
\ref{Lemma2} we obtain that the function $f(\sigma )$ reaches minimum at the
point defined in the Lemma \ref{Lemma2}. The result for this case is summarized below.

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{*}= \sigma_{2} $ and $\min$ $f(\sigma)$= $f(\sigma_{2})$= $C^{BS}(X_{1},\sigma_{2})-C^{mkt}(X_{1})$ & if $\varphi \leq 0$ \\ 
$\sigma ^{*}= \sigma_{2} $ and $\min$ $f(\sigma)$= $f(\sigma_{2})$=$C^{BS}(X_{1},\sigma_{2})-C^{mkt}(X_{1})$& if $\varphi > 0$ and $0< \sqrt{\varphi } < \sigma _{1}$ \\ 
$\sigma ^{*}=\arg \min \left( h(\sigma _{1}),h(\sigma_{2})\right)$ and $\min$ $f(\sigma)$= $\min$ $(f(\sigma_{1}), f(\sigma_{2}))$& if $\varphi > 
0$ and $\sigma_{1} \leq \sqrt{\varphi } \leq \sigma_{2}$ \\ 
$\sigma ^{*}=\sigma _{1}$ and $\min$ $f(\sigma)$= $f(\sigma_{1})$=$C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma_{1})$& if $\varphi > 0$ and $\sqrt{\varphi} > \sigma_{2}$ \\ 
\end{tabular}
\right. 
\]



\end{description}



\begin{itemize}
\item  \underline{Case \#2: $\sigma _{1}=\sigma _{2}$ }- we split the domain
of $\sigma $ into three intervals: $[0,\sigma _{1})\cup [\sigma _{2}]\cup
(\sigma _{2},\infty ]$

\begin{itemize}
\item  For $\sigma $ $\in $ $[0,\sigma _{1})$, the function $f(\sigma )=$ $%
C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )+C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )$.
It is decreasing in $\sigma $ because $\frac{\partial C^{BS}(X,\sigma )}{%
\partial \sigma }$ is positive for every $X$ (see Lemma \ref{Lemma 1}). 
\textit{\ }

\item  For $\sigma =\sigma _{1}=\sigma _{2}$, the function $f(\sigma )=0,$
which is the global minimum of non-negative function $f(\sigma ).$

\item  For $\sigma \in (\sigma _{2},\infty ],$ the function $f(\sigma )=$ $%
-C^{mkt}(X_{1})+C^{BS}(X_{1},\sigma )-C^{mkt}(X_{2})+C^{BS}(X_{2},\sigma )$
is increasing in $\sigma $ because $\frac{\partial C^{BS}(X,\sigma )}{%
\partial \sigma }$ is positive for every $X$ (see Lemma \ref{Lemma 1}).
\end{itemize}


\smallskip Thus, in this case the $\sigma _{1}=\sigma _{2}=\arg \min
f(\sigma )$ such that $f(\sigma _{2})=0$



\item  \underline{Case \#3: $\sigma _{1}$ $>\sigma _{2}$ }- To proceed, we
split the domain for the argument $\sigma $ of the function $f(\sigma )$
into 3 intervals: $[0,\sigma _{2})$ $\cup $ $\left[ \sigma _{2},\sigma
_{1}\right] $ $\cup $ $\left( \sigma _{1},\infty \right) .$

\begin{itemize}
\item  For $\sigma $ $\in $ $[0,\sigma _{2})$, the function $f(\sigma )$ $%
=C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma )+C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma )$
is decreasing in $\sigma $ because $\frac{\partial C^{BS}(X,\sigma )}{%
\partial \sigma }$ is positive for every strike $X$ (see Lemma \ref{Lemma 1}%
)\textit{\ }. Also, $\displaystyle \lim_{\sigma \to \sigma_{2}^{-}}f(\sigma)$=$C^{mkt}(X_{1})-C^{BS}(X_{1}, \sigma_{2})$. Hence $f(\sigma)$ in this interval will be greater than the limit value $C^{mkt}(X_{1})-C^{BS}(X_{1}, \sigma_{2})$.

\item  For $\sigma $ $\in $ $\left[ \sigma _{2},\sigma _{1}\right] $, the $%
f(\sigma )$ $=$ $C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma
)$-$C^{mkt}(X_{2})+C^{BS}(X_{2},\sigma )$ $=$ $C^{mkt}(X_{1})$ $%
-C^{mkt}(X_{2})$ $-$ $h(\sigma )$, where $h(\sigma )=C^{BS}(X_{1},\sigma
)-C^{BS}(X_{2},\sigma ).$ Applying Lemma \ref{Lemma3} we find that the
arg minimum of $f(\sigma )$ corresponds to the argmax of $h(\sigma )$ as
defined and explained in Lemma \ref{Lemma3}.

\item  For $\sigma $ $\in \left( \sigma _{1},\infty \right) $, the function $%
f(\sigma )=$ $-C^{mkt}(X_{1})+C^{BS}(X_{1},\sigma
)-C^{mkt}(X_{2})+C^{BS}(X_{2},\sigma )$ is increasing in $\sigma $ because $%
\frac{\partial C^{BS}(X,\sigma )}{\partial \sigma }$ is positive for every $X
$ (see Lemma \ref{Lemma 1}). \textit{\ }.  Also, $\displaystyle \lim_{\sigma \to \sigma_{1}^{+}}f(\sigma)$=$-C^{mkt}(X_{2})+C^{BS}(X_{2}, \sigma_{1})$. Hence $f(\sigma)$ in this interval will be greater than the limit value $-C^{mkt}(X_{2})+C^{BS}(X_{2}, \sigma_{1})$.
\end{itemize}

The behavior of $f(\sigma)$ when $\sigma_{1}>\sigma_{2}$ is summarized in the table below:
\begin{table}[ht]
\caption{Examining $f(\sigma)$ in the case $\sigma_{1}>\sigma_{2}$} % title of Table
\centering % used for centering table
\begin{tabular}{c c c c} % centered columns (4 columns)
\hline\hline %inserts double horizontal lines
	& $0 < \sigma < \sigma_{2}$& $\sigma_{2} \leq \sigma \leq \sigma_{1}$ & $\sigma > \sigma_{1}$ \\ [0.5ex] 
\hline % inserts single horizontal line
	$f(\sigma)$ & Decreasing & Depends on behavior of $h(\sigma)$ & Increasing \\
\hline % inserts single horizontal line
	$\min f(\sigma)$ & Greater than $f(\sigma_{2})$ & Is either $f(\sigma_{1})$, $f(\sigma_{2}$, or $f(\varphi)$ & Greater than $f(\sigma_{1})$\\
\hline
\end{tabular}
\label{table1} % is used to refer this table in the text
\end{table}

\begin{description}
\item  The function $f(\sigma )$ is decreasing on $[0,\sigma _{2})$ and
increasing on $\left( \sigma _{1},\infty \right) $, hence its argmin is in
the $\left[ \sigma _{2},\sigma _{1}\right] $ interval. Applying the Lemma 
\ref{Lemma3} we obtain that the function $f(\sigma )$ reaches minimum at the
point where $h(\sigma)$ reaches its maximum. This point is specifed in the Lemma \ref{Lemma3}. The result for this case is summarized below.

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{*}= \sigma_{2} $ and $\min$ $f(\sigma)$= $f(\sigma_{2})$=$C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma_{2})$ & if $\varphi \leq 0$ \\ 
$\sigma ^{*}= \sigma_{2} $ and $\min$ $f(\sigma)$= $f(\sigma_{2})$=$C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma_{2})$ & if $\varphi > 0$ and $0< \sqrt{\varphi } < \sigma _{2}$ \\ 
$\sigma ^{*}=\sqrt{\varphi}$ and $\min$ $f(\sigma)$= $f(\sqrt{\varphi})$& if $\varphi > 0$ and $\sigma_{2} \leq \sqrt{\varphi } \leq \sigma_{1}$ \\ 
$\sigma ^{*}=\sigma _{1}$ and $\min$ $f(\sigma)$= $f(\sigma_{1})$=$C^{BS}(X_{2},\sigma_{1})-C^{mkt}(X_{2})$ & if $\varphi > 0$ and $\sqrt{\varphi} > \sigma_{1}$ \\ %
\end{tabular}
\right. 
\]

\end{description}

\end{itemize}

The results in the three cases are summarized as shown in the statement of Theorem \ref{Theorem 1}. 
\newline


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{Theorem \ref{Theorem 2}.}
The argument of the minimax statistic is reached at a  unique assumed volatility $\sigma ^{mm}$ within the interval $\left[ \sigma
_{1},\sigma _{2}\right] $ such that the two dollar errors at this assumed volatility are equal, that is $|C^{mkt}(X_{1})-C^{BS}(X_{1},\sigma ^{mm})|$ $=$ $|C^{mkt}(X_{2})$ - $C^{BS}(X_{2},\sigma ^{mm})|$.

\subparagraph{Proof of theorem \ref{Theorem 2}.} As in Theorem \ref{Theorem 1}, we also have to look at three cases: $\sigma_{1}<\sigma_{2}$, $\sigma_{1}=\sigma_{2}$, $\sigma_{1}>\sigma_{2}$. 

\underline{Case \#1: $\sigma _{1}$ $<\sigma _{2}$ }
\begin{itemize}
\item First, we examine the interval $[\sigma _{1},\sigma _{2}]$. Within this
interval, we show that there is a point $\sigma ^{mm}$ where the two dollar
errors equal. \newline

Introduce a new function  $g(\sigma )$ $\equiv $ $D(X_{1},\sigma
)-D(X_{2},\sigma )$. For $\sigma $ $\in $ $[\sigma _{1},\sigma _{2}]$, we
have: 
\[
g(\sigma _{1})=D(X_{1},\sigma _{1})-D(X_{2},\sigma
_{1})=0-[C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma _{1})]=C^{BS}(X_{2},\sigma
_{1})-C^{mkt}(X_{2})<0
\] 
\[
g(\sigma _{2})=D(X_{1},\sigma _{2})-D(X_{2},\sigma
_{2})=[C^{BS}(X_{1},\sigma _{2})-C^{mkt}(X_{1})]-0=C^{BS}(X_{1},\sigma
_{2})-C^{mkt}(X_{1})>0
\]

It can be inferred from Lemma \ref{Lemma 1} that in this closed interval, $D(X_{1},\sigma _{1})$ = $C^{BS}(X_{1},\sigma)-C^{mkt}(X_{1})$ is a monotonically increasing function  and $D(X_{2},\sigma _{2})$ = $C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma)$ is a monotonically decreasing function, both with respect to $\sigma$. Hence $g(\sigma)$ is monotonically increasing within the interval. Using the Intermediate Value Theorem (see ~\cite{InterValueTheorem}), we can see that there must be an volatility within the interval $[\sigma _{1},\sigma _{2}]$ where $g(\sigma )$ equals 0. Furthermore, this point is unique as $g(\sigma )$ is monotonically increasing. This volatility is the point where the two dollar errors $D(X_{1})$ and $D(X_{2})$ equal. This assumed volatility is the $\sigma ^{mm}$ and the dollar error at that point is denoted $D^{mm}$.

\item Now we prove that $D^{mm}$ is also the minimum of the greater of the
two dollar errors (minimax statistic) on the whole domain of $\sigma $. 
\newline
Because for $\sigma $ $>$ $\sigma _{1}$, $D(X_{1},\sigma )$ = $C^{BS}(X_{1},\sigma)-C^{mkt}(X_{1})$ is a monotonically increasing function (inferred from Lemma \ref{Lemma 1}), $D(X_{1},\sigma )$ for $\sigma $ $>\sigma^{mm}$ is also a monotonically increasing function ($\sigma _{1}<\sigma ^{mm} $). From that we have: 

\[
D^{mm}=D(X_{1},\sigma ^{mm})\leq D(X_{1},\sigma)_{\sigma \geq \sigma ^{mm}}\leq
max[D(X_{1},\sigma),D(X_{2},\sigma)]_{\sigma \geq \sigma ^{mm}}
\]

Because for $\sigma $ $<$ $\sigma _{2}$, $D(X_{2},\sigma )$= $C^{mkt}(X_{2})-C^{BS}(X_{2},\sigma)$ is a monotonically decreasing function (inferred from Lemma \ref{Lemma 1}), $D(X_{2},\sigma )$ for $\sigma $ $<$ $\sigma ^{mm}$ is also a monotonically increasing function ($\sigma
_{2}>\sigma ^{mm}$). From that we have: 

\[
D^{mm}=D(X_{2},\sigma ^{mm})\leq D(X_{2},\sigma)_{\sigma \leq \sigma ^{mm}}\leq
max[D(X_{1},\sigma),D(X_{2},\sigma)]_{\sigma \leq \sigma ^{mm}}
\]

Also, at the point $\sigma^{mm}$, $max[D(X_{1},\sigma),D(X_{2},\sigma)]$ = $D^{mm}$. We can see that $D^{mm}$ is a value of the function $\max_{X \in \{X_{1},X_{2}\}} D(X,\sigma)$ that is less than the maximum at all other points in the domain of $\sigma$. Hence, $D^{mm}$ is the minimax statistic.
\end{itemize}


\underline{Case \#2: $\sigma _{1}$ = $\sigma _{2}$ }\newline
In this case, we show that the point where the minimax statistic occurs is $%
\sigma ^{mm}$ = $\sigma _{1}$=$\sigma _{2}$. The maximum of the two dollar errors at this point is $max[D(X_{1},\sigma ^{mm},D(X_{2},\sigma ^{mm}]$= $max(0,0)$ = 0. As at other points of $\sigma $, both dollar errors are greater than 0, the $\sigma ^{mm}$ described above is the minimum of the maximum of two dollar errors. This means that the minimax statistic occurs at a unique $\sigma $ within $[\sigma _{1},\sigma_{2}]$ that makes the two dollar equal.


\underline{Case \#3: $\sigma_{1}$ \textgreater $\sigma_{2}$ }\newline
\begin{itemize}
\item First, we examine the interval $[\sigma_{2}, \sigma_{1}]$.
Within this interval, we show that there is a point $\sigma^{mm}$ where the
two dollar errors equal. \newline

Let $g(\sigma)$ = $D(X_{1},\sigma)- D(X_{2},\sigma)$. For $\sigma$ $\in$ $%
[\sigma_{2},\sigma_{1}]$, we have: 

\begin{align}
g(\sigma_{2})= D(X_{1}, \sigma_{2}) - D(X_{2}, \sigma_{2}) = [C^{mkt}(X_{1})
- C^{BS}(X_{1},\sigma_{2})]-0=C^{mkt}(X_{1}) - C^{BS}(X_{1},\sigma_{2}) %
\textgreater 0  \nonumber
\end{align}
\begin{align}
g(\sigma_{1})= D(X_{1}, \sigma_{1}) - D(X_{2}, \sigma_{1})
=0-[C^{BS}(X_{2},\sigma_{1})-C^{mkt}(X_{2})
]=-C^{BS}(X_{2},\sigma_{1})+C^{mkt}(X_{2}) \textless 0  \nonumber
\end{align}

It can be inferred from Lemma \ref{Lemma 1} that in this closed interval, $D(X_{1},\sigma _{1})$ = $C^{mkt}(X_{1})- C^{BS}(X_{1},\sigma)$ is a monotonically decreasing function  and $D(X_{2},\sigma _{2})$ = $C^{BS}(X_{2},\sigma)-C^{mkt}(X_{2})$ is a monotonically increasing function, both with respect to $\sigma$. Hence $g(\sigma)$ is monotonically decreasing within the interval.  Using the Intermediate Value Theorem, we can see that there must be a volatility within the interval $[\sigma_{2},\sigma_{1}]$ where $g(\sigma)$ equals 0. Furthermore, this point is unique as $g(\sigma)$ is
monotonically decreasing. It is also the volatility that makes the two dollar errors $D(X_{1},\sigma)$ and $D(X_{2},\sigma)$ equal. This assumed volatility is the $\sigma ^{mm}$ and the dollar error at that point is denoted $D^{mm}$.


\item Now we prove that $D^{mm}$ is also the minimum of the greater of the
two dollar errors (minimax statistic) on the whole domain of $\sigma$.\newline

Because for $\sigma < \sigma^{mm}$, $D(X_{1},\sigma)$= $C^{mkt}(X_{1})- C^{BS}(X_{1},\sigma)$ is a monotonically decreasing function (inferred from Lemma \ref{Lemma 1}), $D(X_{1},\sigma)$  is also a monotonically decreasing function ($\sigma_{1} > \sigma^{mm}$). Thus we have: 
\begin{align}
D^{mm}= D(X_{1},\sigma^{mm}) \leq D(X_{1},\sigma)_{\sigma \leq \sigma^{mm}} \leq
max[D(X_{1},\sigma),D(X_{2},\sigma)]_{\sigma \leq \sigma^{mm}}  \nonumber
\end{align}

Because for $\sigma > \sigma^{mm}$ $D(X_{2},\sigma)$ = $C^{BS}(X_{2},\sigma)-C^{mkt}(X_{2})$ is a monotonically increasing function (inferred from Lemma \ref{Lemma 1}), $D(X_{2},\sigma)$  is also a monotonically increasing function ($\sigma_{2} < \sigma^{mm}$). Thus we have: 
\begin{align}
D^{mm}= D(X_{2},\sigma^{mm}) \leq D(X_{2},\sigma)_{\sigma \geq \sigma^{mm}} \leq max[D(X_{1},\sigma),D(X_{2},\sigma)]_{\sigma \geq \sigma^{mm}}  \nonumber
\end{align}

Also, at the point $\sigma^{mm}$, $max[D(X_{1},\sigma),D(X_{2},\sigma)]$ = $D^{mm}$. We can see that $D^{mm}$ is a value of the function $\max_{X \in \{X_{1},X_{2}\}} D(X,\sigma)$ that is less than the maximum at all other points in the domain of $\sigma$. Hence, $D^{mm}$ is the minimax statistic.

For all cases, we have the same conclusion that the minimax statistic occurs
at the point that makes the two dollar errors equal. Furthermore, this point
is unique in the interval $\sigma_{1}$ and $\sigma_{2}$.
\end{itemize}


\subsection{Auxiliary Lemmas}

\label{sec:AuxiliaryLemmas}In the proof of theorem \ref{Theorem 1} we use
the fact that the Vega (derivative of Black-Scholes call and put option
prices with respect to volatility) is positive. The following lemma explains
this property.

\begin{lemma}
\label{Lemma 1} The Black-Scholes formula for call and put options is a
monotonically increasing function of volatility. \newline
\end{lemma}

The Black-Scholes call option price is $C^{BS}(X,\sigma
)=S_{t}N(d_{1})-Xe^{-r\Delta t}N(d_{2}),$ where $d_{1}$ $=$ $\frac{ln\frac{%
S_{t}}{X}+r\Delta t}{\sqrt{\Delta t}}\frac{1}{\sigma }+\frac{1}{2}\sqrt{%
\Delta t}\sigma $, $d_{2}$ $=$ $\frac{ln\frac{S_{t}}{X}+r\Delta t}{\sqrt{%
\Delta t}}\frac{1}{\sigma }-\frac{1}{2}\sqrt{\Delta t}\sigma ,$ and $N(d_{1})
$ $=$ $\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{d_{1}}e^{-\frac{1}{2}x^{2}dx}.$
Hence $\frac{\partial C^{BS}(\sigma )}{\partial \sigma }$ $=$ $S_{t}\frac{%
\partial N[d_{1}(\sigma )]}{\partial \sigma }-Xe^{-r\Delta t}\frac{\partial
N[d_{2}(\sigma )]}{\partial \sigma }.$ To evaluate this expression we find
that $\frac{\partial N(d_{1}(\sigma ))}{\partial \sigma }$ $=$ $[\frac{-(ln%
\frac{S_{t}}{X}+r\Delta t)}{\sqrt{\Delta t}}\frac{1}{\sigma ^{2}}+\frac{1}{2}%
\sqrt{\Delta t}]$ $\cdot \frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}[d_{1}(\sigma )%
]^{2}}$, and $\frac{\partial N[d_{2,X}(\sigma )]}{\partial \sigma }$ $=$ $[%
\frac{-(ln\frac{S_{t}}{X}+r\Delta t)}{\sqrt{\Delta t}}\frac{1}{\sigma ^{2}}-%
\frac{1}{2}\sqrt{\Delta t}]\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}[%
d_{2}(\sigma )]^{2}}.$ It implies that$\frac{\partial C^{BS}(\sigma )}{%
\partial \sigma }$ $=$ $\frac{-(ln\frac{S_{t}}{X}+r\Delta t)}{\sqrt{\Delta t}%
}\frac{1}{\sigma ^{2}}$ $\frac{1}{\sqrt{2\pi }}$ $\{S_{t}e^{-\frac{1}{2}[%
d_{1}(\sigma )]^{2}}-Xe^{-r\Delta t}e^{-\frac{1}{2}[d_{2}(\sigma )]^{2}}\}$ $%
+$ $\frac{\sqrt{\Delta t}}{2\sqrt{2\pi }}$ $\cdot \{S_{t}e^{-\frac{1}{2}[%
d_{1}(\sigma )]^{2}}+Xe^{-r\Delta t}e^{-\frac{1}{2}[d_{2}(\sigma )]^{2}}\}.$
Now we prove that the terms $S_{t}e^{-\frac{1}{2}[d_{1}(\sigma )%
]^{2}}-Xe^{-r\Delta t}e^{-\frac{1}{2}[d_{2}(\sigma )]^{2}}$ equals zero.
Indeed, after replacing $d_{1}$ by $d_{2}+\sigma \sqrt{\Delta t}$, the
equation $e^{-\frac{1}{2}d_{1}^{2}}$ $=$ $\frac{X}{S_{t}}e^{-r\Delta t}e^{-%
\frac{1}{2}d_{2}^{2}}$ becomes $e^{-\frac{1}{2}(d_{2}^{2}+2d_{2}\sigma \sqrt{%
\Delta t}+\sigma ^{2}\Delta t)}$ $=$ $\frac{X}{S_{t}}e^{-r\Delta t}e^{-\frac{%
1}{2}d_{2}^{2}}.$ Simplification yields $ln[e^{-\frac{1}{2}(2d_{2}\sigma 
\sqrt{\Delta t}+\sigma ^{2}\Delta t)}]$ $=$ $ln[\frac{X}{S_{t}}e^{-r\Delta
t}].$ Plugging the expression for $d_{2}$ gives $-\frac{ln\frac{S_{t}}{X}+(r-%
\frac{1}{2}\sigma ^{2})\Delta t}{\sigma \sqrt{\Delta t}}\sigma \sqrt{\Delta t%
}$ $-$ $\frac{1}{2}\sigma ^{2}\Delta t$ $=$ $ln\frac{X}{S_{t}}-r\Delta t.$
The equality of two sides can be easily verified. Therefore, two identities
follows $\frac{\partial C^{BS}(\sigma )}{\partial \sigma }$ $=$ $\frac{\sqrt{%
\Delta t}}{2\sqrt{2\pi }}\cdot $ $\{S_{t}e^{-\frac{1}{2}[d_{1}(\sigma )%
]^{2}}+Xe^{-r\Delta t}e^{-\frac{1}{2}[d_{2}(\sigma )]^{2}}\}$ and $S_{t}e^{-%
\frac{1}{2}d_{1}^{2}}+Xe^{-r\Delta t}e^{-\frac{1}{2}d_{2}^{2}}=2S_{t}e^{-%
\frac{1}{2}d_{1}^{2}}$. Hence, $\frac{\partial C^{BS}(\sigma )}{\partial
\sigma }$ $=$ $\frac{\sqrt{\Delta t}}{\sqrt{2\pi }}S_{t}e^{-\frac{1}{2}[%
d_{1}(\sigma )]^{2}},$ which is always positive. Thus, the Black-Scholes
formula for a call option is monotonically increasing with respect to
volatility. Application of the put-call parity yield similar result for the
put option.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion

%EndExpansion


\begin{lemma}
\label{Lemma2}Assume that $X_{1}<X_{2}.$ Examine the function $h(\sigma
)=C^{BS}(X_{1},\sigma )-C^{BS}(X_{2},\sigma ),$defined on the interval $%
\left[ \sigma _{1},\sigma _{2}\right] ,$ where $\sigma _{1}<\sigma _{2}$, ( $%
\sigma _{i}$ is implied volatility of the option with strike $X_{i},$ $%
C^{BS}(X_{i},\sigma )$)$.$ Let us denote $\sigma ^{*}\equiv \arg \min $ $%
h(\sigma )$ and $\varphi \equiv \frac{1}{\Delta t}\ln \frac{X_{1}X_{2}}{%
S_{t}^{2}}-2r$ then 

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{*}= \sigma_{2} $ & if $\varphi \leq 0$ \\ 
$\sigma ^{*}= \sigma_{2} $ & if $\varphi > 0$ and $0< \sqrt{\varphi } < \sigma _{1}$ \\ 
$\sigma ^{*}=\arg \min \left( h(\sigma _{1}),h(\sigma_{2})\right)$ & if $\varphi > 0$ and $\sigma_{1} \leq \sqrt{\varphi } \leq \sigma_{2}$ \\ 
$\sigma ^{*}=\sigma _{1}$ & if $\varphi > 0$ and $\sqrt{\varphi} > \sigma_{2}$ \\ %
\end{tabular}
\right. 
\]

%TCIMACRO{\TeXButton{Proof}{\proof}}
%BeginExpansion
%\proof%
%EndExpansion

Note: $\arg \min \left( h(\sigma _{1}),h(\sigma_{2})\right) $ equals either $\sigma _{1}$ or $\sigma_{2}.$
\end{lemma}

We begin by evaluating the derivative of $h(\sigma)$ with respect to $\sigma$. Using Lemma 1, we have: $\frac {\partial h(\sigma)}{\partial \sigma}$ = $\frac {\partial( C^{BS}(X_{1},\sigma)-C^{BS}(X_{2},\sigma ))}{\partial \sigma}$=$\frac{\sqrt{\Delta t}}{\sqrt{2\pi }}S_{t} \{ e^{-\frac{1}{2}[d_{1}(X_{1}, \sigma )]^{2}} - e^{-\frac{1}{2}[d_{1}(X_{2}, \sigma )]^{2}}\}.$ To evaluate $\frac {\partial h(\sigma)}{\partial \sigma}$ we need to compare $d_{1}(X_{1}, \sigma )^{2}$ and $d_{1}(X_{2}, \sigma )^{2}$. Now $d_{1}(X_{1}, \sigma )^{2}-d_{1}(X_{2}, \sigma )^{2}$
=$[d_{1}(X_{1},\sigma)+d_{1}(X_{2},\sigma)][d_{1}(X_{1},\sigma)-d_{1}(X_{2},\sigma)]$=$\frac{ln \frac{S_{t}}{X_{1}}+ln \frac{S_{t}}{X_{2}} +(2r+ \sigma^{2}) \Delta t}{\sigma \sqrt{\Delta t}}  \frac{ln \frac{S_{t}}{X_{1}}-ln \frac{S_{t}}{X_{2}}}{\sigma \sqrt{\Delta t}}$=$[\sigma^{2} - (\frac{1}{\Delta t}ln \frac{X_{1} X_{2}}{S_{t}^{2}}-2r)] \frac{ln \frac{X_{2}}{X_{1}}}{\sigma ^{2}} $ = $(\sigma^{2} - \varphi) \frac{ln \frac{X_{2}}{X_{1}}}{\sigma ^{2}} $. Since $X_{1} < X_{2}$, the term $\frac{ln \frac{X_{2}}{X_{1}}}{\sigma ^{2}} $ is positive, the sign of $\frac {\partial h(\sigma)}{\partial \sigma}$ depends on the equality between $\sigma^{2}$ and $\varphi$. \newline


We examine two different alternatives $\varphi >0$ and $\varphi \leq 0.$


\begin{itemize}
\item   Consider first the $\varphi >0$ case, that is $S_{t}$ $<$ $\sqrt{X_{1}X_{2}}e^{-r\Delta t}$ or $\frac{1}{\Delta t}\ln \frac{S_{t}^{2}}{%
X_{1}X_{2}}+$ $2r<0$. The following table evaluates the derivative $\frac {\partial h(\sigma)}{\partial \sigma}$ and the behavior of $h(\sigma)$ on its domain $(0, +\infty)$.



\begin{table}[ht]
\caption{Examining $\frac {\partial h(\sigma)}{\partial \sigma}$ for intervals of $\sigma$} % title of Table
\centering % used for centering table
\begin{tabular}{c c c c} % centered columns (4 columns)
\hline\hline %inserts double horizontal lines
	& $0 < \sigma < \sqrt{\varphi}$& $\sigma = \sqrt{\varphi}$ & $\sigma > \sqrt{\varphi}$ \\ [0.5ex] 
\hline % inserts single horizontal line
	$d_{1}(X_{1}, \sigma)^{2} - d_{1}(X_{2}, \sigma)^{2}$ & Negative & 0 & Positive \\
\hline
	$\frac {\partial h(\sigma)}{\partial \sigma}$ &Positive  & 0 &Negative \\ 
\hline 
	$h(\sigma)$ & Increasing &$h(\sqrt{\varphi})$ & Decreasing\\ 
\hline
\end{tabular}
\label{table1} % is used to refer this table in the text
\end{table}

Based on the table we conclude the following for the global minimum of $h(\sigma)$ on the interval $[\sigma_{1}, \sigma_{2}]$.
\begin{itemize}
\item
If $\sqrt{\varphi} < \sigma_{1} < \sigma_{2}$ then the minimum is at $\sigma^{*}$ = $\sigma_{2}$ since $h(\sigma)$ decreases in the interval $[\sigma_{1}, \sigma_{2}]$.
\item
If $ \sigma_{1} \leq \sqrt{\varphi} \leq \sigma_{2}$ then the minimum is at $\sigma ^{*}=\arg \min \left( h(\sigma _{1}),h(\sigma_{2})\right)$ since $h(\sigma)$ increases in the interval $[\sigma_{1}, \sqrt{\varphi}]$ then decreases in the interval $[\sqrt{\varphi}, \sigma_{2}]$.
\item
If $\sigma_{1} < \sigma_{2}<\sqrt{\varphi}$ then the minimum is at $\sigma^{*}$ = $\sigma_{1}$ since $h(\sigma)$ increases in the interval $[\sigma_{1}, \sigma_{2}]$.
\end{itemize}


\item  Now examine the alternative $\varphi \leq 0$ case, that is $S_{t} \leq \sqrt{X_{1}X_{2}}\exp \left[ -r\Delta t\right] $ or $\frac{1}{\Delta t}\ln 
\frac{S_{t}^{2}}{X_{1}X_{2}}+$ $2r>0.$ In this case for all $\sigma ^{2}$ we have $\frac{\partial h(\sigma )}{\partial \sigma }<0$ because $(\sigma^{2} - \varphi) \frac{ln \frac{X_{2}}{X_{1}}}{\sigma ^{2}} >0$ and hence $d_{1}(X_{1}, \sigma)^{2} > d_{1}(X_{2}, \sigma)^{2}$. Since $h(\sigma )$ is decreasing on the domain $\left[ \sigma _{1},\sigma _{2}\right]$, $\sigma_{2}=\arg \min h(\sigma ).$%


\end{itemize}
The lemma is proved with the summarized results shown in the statement.


\begin{lemma}
\label{Lemma3}Assume that $X_{1}<X_{2}.$ Examine the function $h(\sigma
)=C^{BS}(X_{1},\sigma )-C^{BS}(X_{2},\sigma ),$defined on the interval $%
\left[ \sigma _{1},\sigma _{2}\right] ,$ where $\sigma _{1} > \sigma _{2}$, ( $%
\sigma _{i}$ is implied volatility of the option with strike $X_{i},$ $%
C^{BS}(X_{i},\sigma )$)$.$ Let us denote $\sigma ^{**}\equiv \arg \max $ $%
h(\sigma )$ and $\varphi \equiv \frac{1}{\Delta t}\ln \frac{X_{1}X_{2}}{%
S_{t}^{2}}-2r$ then 

\[
\left\{ 
\begin{tabular}{ll}
$\sigma ^{**}= \sigma_{2} $ & if $\varphi \leq 0$ \\ 
$\sigma ^{**}= \sigma_{2} $ & if $\varphi > 0$ and $0< \sqrt{\varphi } < \sigma _{2}$ \\ 
$\sigma ^{**}=\sqrt{\varphi}$ & if $\varphi > 0$ and $\sigma_{2} \leq \sqrt{\varphi } \leq \sigma_{1}$ \\ 
$\sigma ^{**}=\sigma _{1}$ & if $\varphi > 0$ and $\sqrt{\varphi} > \sigma_{1}$ \\ %
\end{tabular}
\right. 
\]

\end{lemma}

\begin{itemize}
\item
In the case when $\varphi \leq 0$, using Lemma \ref{Lemma2} we have $h(\sigma)$ always decreases in the interval $[\sigma_{2}, \sigma_{1}]$. Hence $\max h(\sigma)$ in this interval is reached at $\sigma^{**}$ = $\sigma_{2}$
\item
If $\varphi >0$, similar to the proof of the case $\sigma _{1} < \sigma _{2}$, we take the derivative and examine the behavior of $h(\sigma)$. By comparing the value of $\sigma_{2}$, $\sqrt{\varphi}$ and $\sigma_{1}$, we obtain the following result for the global maximum of $h(\sigma)$ on the interval $[\sigma_{2},\sigma_{1}]$.

\begin{itemize}
\item
If $\sqrt{\varphi} < \sigma_{2} < \sigma_{1}$ then the maximum is at $\sigma^{**}$ = $\sigma_{2}$ since $h(\sigma)$ decreases in the interval $[\sigma_{2}, \sigma_{1}]$.
\item
If $ \sigma_{2} \leq \sqrt{\varphi} \leq \sigma_{1}$ then the maximum is at $\sigma^{**}$=$\sqrt{\varphi}$ since $h(\sigma)$ increases in the interval $[\sigma_{2}, \sqrt{\varphi}]$ then decreases in the interval $[\sqrt{\varphi}, \sigma_{1}]$.
\item
If $\sigma_{2} < \sigma_{1}<\sqrt{\varphi}$ the the maximum is at $\sigma^{**}$ = $\sigma_{1}$ since $h(\sigma)$ increases in the interval $[\sigma_{2}, \sigma_{1}]$.
\end{itemize}
\end{itemize}




\end{document}
